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Chapter 20 Section 2 self-ionization - the reaction in which two water molecules produce ions H 2 O + H 2 O H 3 0 + + OH - hydroxide ion (OH - ) – what a water molecule that loses a hydrogen ion becomes hydronium ion (H 3 0 + ) – what a water molecule that gains a hydrogen ion becomes

Chapter 20 Section 2

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Chapter 20 Section 2. self-ionization - the reaction in which two water molecules produce ions H 2 O + H 2 O  H 3 0 + + OH - hydroxide ion (OH - ) – what a water molecule that loses a hydrogen ion becomes - PowerPoint PPT Presentation

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Page 1: Chapter 20 Section 2

Chapter 20 Section 2

self-ionization- the reaction in which two water molecules produce ions

H2O + H2O → H30+ + OH-

hydroxide ion (OH-) – what a water molecule that loses a hydrogen ion becomes

hydronium ion (H30+) – what a water molecule that gains a hydrogen ion becomes

Page 2: Chapter 20 Section 2

-in pure water at 25°C the concentration of [H+] and [OH-] are 1.00 x 10-7mol/L

* [H+] = [OH-] in pure waterneutral solution- [H+] and [OH-] are equal

-in any aqueous solution, as [H+] inc. [OH-] dec. and vice versa

-in aqueous solutions: [H+] x [OH-] = 1.00 x 10-14M2

Page 3: Chapter 20 Section 2

Kw = ion product constant for water Kw = [H+] x [OH-] = 1.0 x 10-14 M2

[H+] = Kw / [OH-]

[OH-] = Kw / [H+]

Page 4: Chapter 20 Section 2

-not all solutions are neutral → some release H+ when dissolved in water

ex- HCℓ → H+ + Cℓ-

-such solutions have a greater [H+] than [OH-] acidic solution- [H+] is greater than [OH-]

-so [H+] > 1.0 x 10-7 M-some solutions release [OH-]ex- NaOH → Na+ + OH-

basic solution- [H+] is less than [OH-]-so [H+] < 1.0 x 10-7 M

**looking at negative exponent (ex -7 < -2)

Page 5: Chapter 20 Section 2

-[H+] can also be expressed using pH scale

pH = potential hydrogen

-pH scale ranges from 0-14-pH < 7 = acids 0= strongly acidic-pH > 7 = bases 14= strongly basic-pH = 7 = neutral solution

**based on scales of 10-page 584 Table 20.2

Page 6: Chapter 20 Section 2

pH = -log [H+]pOH = -log [OH-]pH + pOH = 14

Examples: [H+]= 1.0 x 10-7

pH = -log 1.0 x 10-7

pH = 7.00 neutral solution

[H+]= 4.3 x 10-3

pH = -log 4.3 x 10-3 pH = 2.37 acidic solution

Page 7: Chapter 20 Section 2

• To find [H+] or [OH-] from a pH or pOH use:

[H+] = antilog - pH

[OH-] = antilog –pOH

antilog = 10x button

Ex: pH = 3.40Find [H+][H+] = antilog -3.40

= 3.98 x 10-4M